Tuesday, February 15, 2022

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange


How to understand the intrinsic definition fo the projective tangent space

Posted: 15 Feb 2022 03:56 AM PST

Introduction

I am a physicist struggling with some basic definitions and concepts from algebraic geometry. Therefore I apologize if I make mistakes, I'm just learning! By the way, English is not my mother tongue, sorry if I made grammar or vocabulary mistakes!

Tangent space to a point

Let $X$ be a projective variety ($X\subset \mathbb{P}^n$). The intrinsic definition of the tangent space to $X$ at $p\in X$ is $T_p(X)=(m_p\backslash m_p^2)^\star$, where $m_p$ is the maximum ideal of functions vanishing at p. Equivalently, one can define $T_p(X)= Z(\{d_pf ~|~ f\in I(X) \text{ homogeneous } \})$, where $Z(\{f_i\}_{i\in J\subset \mathbb{N}})=\{a\in \mathbb{P} | f_i(a)=0 \forall i\in J\}$.

Question

Interpretation of $T_p(X)=(m_p\backslash m_p^2)^\star$ : I don't understand the meaning of this definition. I understand, following $T_p(X)= Z(\{d_pf ~|~ f(t)=0 ~\forall t\in X\})$, that the tangent space to a point is the space where directional derivatives vanish, but I can't link this insight to the so-called intrinsic definition.

Even positive k >= 4 for which the linear diophantine equation d(k)*X + (k - d(k))*Y - k*(k+1)/2 = 0 has a solution ? d(k) number of divisors of k.

Posted: 15 Feb 2022 03:55 AM PST

This LDE has allways a solution if k is a prime number. Under k < 10^5 I found only k=20 and k=432 nonprimes which holds. Are there any other even k-s ? This started as a question : For which k-s is the arithmetic mean of divisors of k an integer AND the arithmetic mean of nondivisors of k is an integer ?

Commutator of the wreath product.

Posted: 15 Feb 2022 03:55 AM PST

I am reading a textbook and it mentions the following: Let $A$ be a non-trivial abelian group and let $K$ be a solvable group.

If $G=K\wr A$ then $G'\leq K^A$. It is mentioned that this follows from the fact that $A$ is abelian, but nothing else is specified.

I am wondering if $G'\leq K^A[A,A]=K^A$. But I don't know why this should be true.

Any hint or help would be appreciated.

Dimension of Tensor Product Space should be $\leq dim(V)$ + $dim(W)$

Posted: 15 Feb 2022 03:55 AM PST

Let $V$ and $W$ be two vector spaces of dimensions $m$ and $n$ over the same field $F$.

$V \otimes W$ is defined as the vector space of equivalence classes of vector space($F(V \times W)$) with bases as the tuples of form $(v,w)$ under the relation specified by: $$ \begin{align} (v_1 + v_2, w)&=(v_1, w)+(v_2, w)\\ (v, w_1+w_2)&=(v, w_1)+(v, w_2)\\ (sv,w)&=s(v,w)\\ (v,sw)&=s(v,w) \end{align} $$

Now in the following, I try to construct a representation of each equivalence class.

Let

$$z = a_1(v_1,w_1)+a_2(v_2, w_2)+ ... + a_n(v_n, w_n) \in F(V \times W)$$

Then:

$$ \begin{align} z &\sim a_1(v_1,w_1)+a_2(v_2, w_2)+ ... + a_n(v_s, w_s) \qquad (1)\\ &\sim (a_1v_1,w_1)+(a_2v_2, w_2)+ ... + (a_sv_s, w_s) \qquad (2)\\ &\sim (a_1v_1+a_2v_2+...+a_sv_s, w_1+w_2+...+w_s) \end{align} $$ where, $a_i \in F, v_i \in V, w_i \in W \text{ for } i=1,2, ..., s$.

Now the additions, are in original vector spaces, $V$ and $W$, rather than $F(V\times W)$, so we can just perform the addition and get some element $v$ of $V$ in the first slot and some element $w$ of $W$ in the second slot of the tuple.

So we see that the equivalence class of any $z \in F(V \times W)$ can be represented by $(v,w)$, where $v \in V$ and $w \in W$. But are these representations, unique?

In going from step $(1)$ to $(2)$, each of the $a_i$'s could have been moved in front of either $v_i$ or $w_i$, and the equivalence class would still have been the same. But, the final vectors in the two slots of the tuple would have been different. This suggests there are different representations of the same equivalence class.

In other words, we may not have unique equivalence classes corresponding to each tuple of the form $(v,w)$. So, we let $\{v_1, v_2, ..., v_m\}$ and $\{ w_1, w_2, ..., w_n \}$ be a particular basis of $V$ and $W$, and proceed further:

$$ \begin{align} (v,w) &\sim (\sum_{i} \alpha_i v_i, \sum_{j} \beta_j w_j)\\ &\sim \sum_{i,j} (\alpha_i v_i, \beta_j w_j)\\ &\sim \sum_{i,j} \alpha_i \beta_j (v_i, w_j) \end{align} $$

Previously we showed that all equivalence classes had a representation: $(v,w)$. And now, we have shown that all equivalence classes have a representation of the form $\sum_{i,j} \alpha_i \beta_j (v_i, w_j)$. As the set:

$$\{(v_i, w_j) | i \in [m], j \in [n]\}$$

spans the equivalence classes, and as there are at most $m+n$ variables that we choose here, ($m$ $\alpha_i$'s and $n$ $\beta_j$'s), I don't see how the Tensor product could possibly have dimension as $m \times n$.

Where did I go wrong?

Converting chances to probabilities

Posted: 15 Feb 2022 03:36 AM PST

I have some data from a reference book that is given as chances per million per year. For example, it says the chances of a catastrophic explosion occurring are 2 chances per million per year.

How do I convert this to a probability to use later in some probability calculations?

Thank you

Tree diagram of a chess tournament amongst 4 players

Posted: 15 Feb 2022 03:35 AM PST

I'd like to know which is the tree diagram of a chess tournament among 4 players. In particular, given 4 players A B C D, we know that there are the drawings of the players in order to create the couple of players. The 3 possibilities of drawing are (uniformly distributed so each of them has a probability of $1/3$):

$$ T_1: (A \, versus \, B) (C \, versus \, D) \\ T_2: (A \, versus \, C) (B \, versus \, D) \\ T_3: (A \, versus \, D) (B \, versus \, C) $$

We know also that A is stronger than B, B is stronger than C, and C is stronger than D; in a challenge between 2 players, the stronger player has the win probability of $2/3$, the other player of $1/3$.

The goal is to get the win probability of C. If I consider the first tournament $T_1$ (its drawing probability is $\color{red}{1/3}$), the win probability of C in the first challenge (C vs. D) is $\color{lightgreen}{2/3}$, instead the win probability of C in the second challenge (C vs. A or C vs. B) is $\color{orange}{1/3}$, so:

$$ P(C | T_1) = \color{red}{\frac{1}{3}} \color{lightgreen}{\frac{2}{3}} \color{orange}{\frac{1}{3}} $$

Tree diagram of the above partial situation ($T_1$ tournament): what is the tree diagram?

enter image description here

Some calculations about divergence of vector field on Riemannian manifold

Posted: 15 Feb 2022 03:24 AM PST

Assume first that we have a Riemannian manifold $(M,g)$. Furthermore, $X$ is a vector field on M and $\nabla$ is the Levi-Civita connection as usual. Let $\{e_i\}$ be an orthonormal basis on M.

Then how can we get $\operatorname{div} X =\sum_i\langle e_i,\nabla_{e_i}X\rangle$? Where $\operatorname{div} $ represents the divergence of $X$. In other words, $\operatorname{div} X=\operatorname{tr}(\nabla X)$.

I try to use the definition of covariant differential and we have $$\operatorname{div} X = \sum_{i}\nabla X(e_i,e_i)=\sum_{i}\nabla_{e_i}X(e_i)$$ but what's next? I have found many books and they just ignore the detail so can someone help me ? Many thanks to you.

How to approach the problem of summation of Eisenstein series on shifted lattices?

Posted: 15 Feb 2022 03:13 AM PST

This question is an attempt to complete the issues discussed in a previous question of mine (How did Gauss sum Eisenstein series?), since my updated question did not recieve any attention. Im my previous question, I mentioned that Gauss wrote an infinite series for the logarithm of the denominator of $\mathbb{sinlemn}(z) = \frac{M(z)}{N(z)}$. Since

$$N(z) = \prod (1-\frac{z}{((m+\frac{1}{2})+(n+\frac{1}{2})i)\varpi})$$

(that is, $\mathbb{sinlemn}(z)$ has poles at Gaussian half-integers multiples of $\varpi$). Gauss wrote:

$$\mathbb{log}N(z) =\frac{1}{12}z^4 - \frac{1}{280}z^8 +\frac{1}{4950}z^{12} - ...$$.

Since the logarithm of a infinite product equals an infinite sum of logarithms, one gets that:

$$\mathbb{log}N(z) = \sum \mathbb{log}(1-\frac{z}{((m+\frac{1}{2})+(n+\frac{1}{2})i)\varpi})$$,

and by the taylor series expansion of $\mathbb{log}(1-z)$ one gets that Gauss's infinite series for $\mathbb{log}N(z)$ is equivalent to te summation of a kind of "generalized Eisenstein series" in which the lattice is shifted by $(\frac{1}{2}+\frac{1}{2}i)\varpi$. Since such shifted lattice cannot be generated by some action of the modular group on the lattice of Gaussian integers (if it was, one could use the modularity of the Eisenstein series and deduce the series for $\mathbb{log}N(z)$ from that of $\mathbb{log}M(z)$), I wondered what tools enable to sum such series and what was Gauss's original method in this case. I tried to make a Google search about "modular forms defined on shifted lattices", but without success.

Side remark

I have no intention to "spam" StackExchange Mathematics with multitudes of more or less similar questions, so I have no problem to close this question and instead get an answer to my original (updated) question, if other users will vote to do so.

Traditional Marriage Algorithm (TMA), optimal and pessimal mating

Posted: 15 Feb 2022 03:10 AM PST

In the TMA where boys propose and girls reject can we say that boys get their optimal mate because boys go first and the girls wait.

Let's define ideal mate where ideal mate is the one on top of their list. With each proposal, the boys either strike the topmost girl on his list or wait, after eliminating the top girl, the topmost girl on his list after the elimination is his ideal mate. With each elimination, the boy would still get his ideal mate but girls will not get their ideal mate, because they have to choose from the boys proposing to her. Some girls might get their ideal mate but not all. Whereas every guy will get his ideal mate.

Is my explanation right or is there some logical fallacy?

Prove the shape is a trapezoid

Posted: 15 Feb 2022 03:38 AM PST

Prove the shape is a trapezoid when :

AB + DC = 2EF

AE = ED , BF = FC

image

Prove that $1 + 3^n$ is divisible by $4$ [closed]

Posted: 15 Feb 2022 03:18 AM PST

Prove that $1 + 3^n$ is divisible by $4$, if $n$ is a positive odd integer,
by using First principle of Mathematical Induction.

Closedness of $L^\infty$ with compact support under some constraints from solutions of ODEs

Posted: 15 Feb 2022 03:09 AM PST

Let $u\in L^\infty([0,T];[-1,1]^m)$ and let $x_u\in C_0([0,T];\mathbb{R}^n)$ be the solution of

$$\dot{x}_u(t) = f(x_u(t),u(t)) \textrm{ with } x(0)=x^0$$

where $\exists C < +\infty$ such that $\forall u, \parallel f(x,u)\parallel < C(1+\parallel x\parallel)$. As a consequence (Gronwall lemma) $\exists K\subset \mathbb{R}^n$ a compact set such that $\forall u, \forall t\in[0,T], x_u(t)\in K$.

Let $g_i\in C_2(\mathbb{R}^n;\mathbb{R})$, $i=1,\dots,p$ and let us define two sets $U_0,U_1$

$$U_0 = \lbrace u\in L^\infty([0,T];[-1,1]^m)\textrm{ s.t. } g_i(x_u(t)) > 0 \forall t\in[0,T],\forall i=1,\dots,p \rbrace$$ $$U_1 = \lbrace u\in L^\infty([0,T];[-1,1]^m)\textrm{ s.t. } g_i(x_u(t)) \geq 0 \forall t\in[0,T],\forall i=1,\dots,p \rbrace$$ where we assume that $U_0\neq \emptyset$ and $U_1\neq \emptyset$

Does the following holds:

$$U_1 \subseteq \textrm{clos}(U_0)$$ where clos(.) is the closure of the set.

Conditional Probability of Discrete Independent Events

Posted: 15 Feb 2022 03:09 AM PST

Let $Y$ and $Z$ be discrete, independent random variables. Then

$$P(Y = i | Y < Z) = P(Y = i)$$

Right? Because $Y$ and $Z$ are independent, the fact that $Y < Z$ doesn't tell us anything about $Y$, right?

On the other hand, using the conditional probability formula:

$$P(Y = i | Y < Z) = \frac{P(Y = i \wedge Y < Z)}{P(Y < Z)}$$

$$P(Y = i | Y < Z) = \frac{P(i < Z)}{P(Y < Z)}$$

$$P(Y = i | Y < Z) = \frac{P(i < Z)}{\sum_{0}^{\infty}P(Z > y) \cdot P(Y=y)}$$

Which seems to give a completely different answer that depends on $Z$...

What am I doing wrong here?

A question about normal matrices and length value

Posted: 15 Feb 2022 03:14 AM PST

I was given the following question, and I find it hard to prove.

Let $A \in \Bbb C^{N \times N}$ be normal matrix. Prove that the following equation is correct $$\| (A+I)v \| = \|(A^* + I)v\|$$


My thoughts: If A was unitary, so the equation was correct since A will not change v's length. But this matrix is normal, not unitary.
There is no helpful information about A that lets me guess that A* = A.
So the only piece of information I have is that AA* = (A*)A, and that A is diagonalized under complex numbers. Means it has eigenvalues and eigenvectors.
The only thing I can think about is that MAYBE if x is an eigenvalue for A with eigenvector t, so x is also an eigenvalue for A* with eigenvector t.
But I'm not quite sure about this sentence, so I don't think about using it to solve this question.
Can anyone please hint me to the solution for this problem? Thanks in advance.

How to perform Lagrange interpolation on a 3D mesh?

Posted: 15 Feb 2022 03:20 AM PST

We have a function-like system that produces a output $z$ value for each 2-dimensional $(x,y)$ input. Suppose we have sampled the system using a set of non-uniform $x,y$ inputs (inputs are not in rectangular grids), and then the generated $(x,y) => z$ data is connected into a triangle mesh through some simple approach. How to perform smoothed interpolation on this stuff (given any free $(x,y)$, produce a $z$), like the Lagrange interpolation performed on 1-dimensional $x => y$ data?

And more, how to expand it to a system that has 3D input like $(x,y,z) => w$?

Find the number of the continuous function between this topological spaces

Posted: 15 Feb 2022 03:57 AM PST

Let the $(X,T) = \{ \{1\}, \{1,2\}, \{1,3,4\} ,\{1,2,3,4\}, \{1,2,5\}, X, \phi \}$ on $X = \{1,2,3,4,5\}$

Find the number of the continuous function $f: (X, T) \to (X,T)$ satisfying the $f(2)=2, f(3) =3$

Since the $f(\bar A) \subset \bar f(A)$ for the continuous function $f$, $f(5) \in \{2,5\}$ and $f(4) \in \{3,4\}$ considering the $A =$ $\{2,5\}$ or $\{3,4\}$ each cases. Plus $1$ is the isolated point, $f$ is always continuous at $1$. Therefore $f(1) \in \{1,2,3,4,5\}$. So my answer is $2\times 2\times 5 =20$. But the answer was $4$. It said $f$ is continuous only for the $f(1)=1$. Why the answer is $4$? How can I derive it? Plus, What the point did I have a mistake?

$(n^1+n)/2$ sequence

Posted: 15 Feb 2022 03:01 AM PST

This is going to be hard to explain so I'll just give an example

Let's say we have a standard arithmetic sequence that goes up by 1 each time

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

The last number (in this case, 10) is n

Scenario z:

Pick 2 numbers in the sequence and multiply them. The product should be equal to the rest of the numbers added together. For example, 6 and 7 (let's call 6 and 7 x and y for now) are between 1 and 10 and if multiplied together, we get 42. Add 1+2+3+4+5+8+9+10 (don't count the numbers that were multiplied together) and we also get 42.

n can only be an integer, no decimals involved here.

x and y has to be an integer between 0 and and n, and x cannot be equal to y

The question is, is there a mathematical way for figuring out the different numbers that can represent n that satisfies scenario z? And if n does satisfy the given scenario, is there a way to figure out x and y without having to try every possible combination?

$n^2+1$ numbers seems to always work (10, 17, 26, 37 etc), and I have no clue why.

I know I explained this pretty badly, feel free to ask any questions regarding this.

Relation between $\int_0^{\frac{\pi}4}(1+\tan x)^2dx$ and $\int_0^1\frac1{(1+x)^2(1+x^2)}dx$.

Posted: 15 Feb 2022 02:57 AM PST

Let $\displaystyle I_1=\int_0^{\frac{\pi}{4}}(1+\tan x)^2dx$ and $\displaystyle I_2=\int_0^{1}\frac{1}{(1+x)^2(1+x^2)}dx$ then find $\displaystyle\frac{I_1}{I_2}$

My method-

I was able to solve $I_1$ using standard formulae and got $I_1=1+\ln2$ , similarly I solved for $I_2$ using partial fraction decomposition and got $I_2=0.25 (1+\ln 2)$ and therefore required ratio will be $4$.

Is there some other way to solve this question?

Number of 3-element subsets with only 1 common element

Posted: 15 Feb 2022 03:14 AM PST

In a party, there are $n$ guests attending where $3$ guests can seat around a table. What is the maximum number of ways in which they can be seated so that no $2$ guests sit together more than once?

Method $1$: This can be solved by considering a polygon with $n$ sides and connected diagonals (or a fully-connected graph with $n$ nodes) and determining the max. number of triangles possible with no common edge (or max. no. of $3$-edged cyclic paths such that there are no path overlaps). By doing this, I got the answer $n-3$, albeit not with a definite proof.

Method $2$: It can also be proved by finding all the $3$-element subsets of the set $\{1,2,...,n\}$ such that intersection of any $2$ subsets will give an empty set or $1$-element set.

This is what I figured out as of now. But, I am unable to conclusively get an answer for both the methods. How to prove the results and solve for the original question with both of the above methods? Also, I am interested in knowing other interpretations of the problem other than Methods 1&2.

How do I deal with the integration limits of the integrals that show up when showing that$Kx(t)=\int_0^tK(t,\tau)x(\tau)d\tau, x\in X$ is compact?

Posted: 15 Feb 2022 03:30 AM PST

Let $X=C(I)$ with the sup norm $\|\cdot\|_\infty$, where $I=[0,1]$. Let $K$ be a Volterra integral operator:

$$Kx(t)=\int_0^tK(t,\tau)x(\tau)d\tau, x\in X$$

Show that K: $X\rightarrow X$, is compact

I am having trouble with the $t$ variable present both in $K(t,\cdot)$ and in the integral integration limit. Most examples I've found are easier because they have $K(t,\tau)=1$ or the integral is from 0 to 1.

I have already shown that K is linear,

1)K is bounded:

$|Kf(t)|=|\int_0^tK(t,\tau)f(\tau)d\tau|\le \int_0^t|K(t,\tau)f(\tau)|d\tau \le \int_0^1|K(t,\tau)f(\tau)|d\tau \le\| f\|_{\infty}\int_0^1|K(t,\tau)|d\tau $

Since $K$ is continuous over a compact set, by Weierstrass theorem $\|K(\cdot, \cdot)\|\le M$, with $M$ some real number so $|Kf(t)|=\| f(\tau)\|_{\infty}M $

Then $\|Kf\|_\infty=\sup_t|Kf(t)|\le\| f\|_{\infty}M $ so K is bounded and $\|K\|\le M $

2)$K$ is compact

To show T is compact I have to use Ascoli-Arzelà theorem to the image of the unit ball with rispect to the sup norm: $K(B(0;1))$, so I need to verify the hypotheses of the theorem

$B=B(0;1)=\{f \in C[0,1], \|f\|_\infty <1\}$

i) K(B) is bounded because $\|Kf\|_\infty=\sup_t|Kf(t)|\le\| f\|_{\infty}M \le M \forall f \in B$

ii) K(B) is equicontinuous:

From the uniform continuity of $K$ with respect to the first variable, I have that $\forall \varepsilon >0, \exists \delta > 0$ such that $\forall t_1, t_2 \in [0,1]$, $|K(t_1,\cdot)-K(t_2, \cdot)|\le \varepsilon$...(*)

$|\int_0^tK(t,\tau)f(\tau)d\tau -\int_0^{t_0}K(t_0,\tau)f(\tau)d\tau|$

I don't know how to bound this, if I use the triangle inequality I'd lose the minus sign and then I can't use (*). Can someone shed some light? Is the rest of the proof ok?

hint: The book says that there is a discontinuity line at $\tau = t$ and that in the analogous easier proof with upper integration limit =1 that is given in the book, the continuity of K on the points $(t,\tau)$ for $t<\tau$ was not used. I don't know what they mean or how to use this hint

Edit This is the proof they give in the book for the easier case. Somehow the hint is suggesting that either the proof is the same or there is a slight modification, but I can't figure it out

enter image description here

enter image description here

Use of the phrase "tangent vector of a curve"

Posted: 15 Feb 2022 03:32 AM PST

Let us understand a curve as a differentiable map $f : J \to \mathbb R^n$ defined on an open interval $J \subset \mathbb R$. The derivative $f'(p)$ of $f$ at $t_0 \in J$ is given as the vector $(f'_1(t_0),\ldots,f'_n(t_0)) \in \mathbb R^n$ where the $f_i : J \to \mathbb R$ are the coordinate functions of $f$.

I think for $n > 1$ it is usual to say that $f'(p)$ is the tangent vector to the curve $f$ at the point $t_0$ or the velocity vector of the curve $f$ at the point $t_0$.

For $n = 1$ one can find the wording velocity vector, but I have never seen that $f'(t_0)$ is called the tangent vector of $f :J \to \mathbb R$ at $t_0$. The definition of the derivative $f'(t_0)$ as the limit $\lim_{t \to t_0}\dfrac{f(t)-f(t_0)}{t-t_0}$ is nevertheless motivated by the concept of tangent by saying that $f'(t_0)$ is the slope of the tangent of the graph $G(f) = \{(t,f(t)) \mid t \in J \} \subset \mathbb R^2$ at the point $(t_0,f(t_0))$. There is also a notational relation to the case $n > 1$: If we consider the curve $\bar f : J \to \mathbb R^2, \bar f(t) = (t,f(t))$, then we get $f'(t_0)$ as the second coordinate of the tangent vector $\bar f'(t_0) \in \mathbb R^2$.

Finally, if we consider a smooth ($C^\infty$) curve $f$ and a point $t_0 \in J$ such that $f'(t_0) \ne 0$, then $M = f(J)$ is locally around $p_0 = f(t_0)$ a smooth one-dimensional submanifold of $\mathbb R^n$. It has a tangent space $T_{p_0}M$ at $p_0$ which we may regard as a one-dimensional linear subspace of $\mathbb R^n$ and all $v \in T_{p_0}M$ are called tangent vectors at $M$ at $p_0$. This suggests that all scalar multiples of the tangent vector $f'(t_0)$ can also be regarded as tangent vectors (which appears reasonable to me).

I find this notationally confusing. The word "tangent vector" seems to have various different interpretations, but in the most elementary case $n = 1$ it is not used. Would it be better to avoid using the name "tangent vector" for $f'(t_0)$, but to use the unambiguous "velocity vector"? Perhaps somebody can help me to clarify my disorientation.

$L^2$-norm of Hilbert space-valued functions

Posted: 15 Feb 2022 03:07 AM PST

Let $\mathcal{H}$ with scalar product $\langle\cdot,\cdot\rangle$ and induced norm $||\cdot||_{\mathcal{H}}$ be a Hilbert space. The Hilbert space $L^2(\mathbb{R}^d,\mathcal{H})$ of Hilbert space-valued square-integrable functions is defined to be the space of all measurable $u:\mathbb{R}^d \to \mathcal{H}$ such that $$ ||u||^2_{L^2} = \int_{\mathbb{R}^d} ||u(x)||_\mathcal{H}^2 \mathrm{d}x < \infty. $$ A similar (but weaker) norm is given by $$ ||u||_X^2 = \sup_{||\psi||_{\mathcal{H}}=1} \int_{\mathbb{R}^d} |\langle\psi,u(x)\rangle|^2 \mathrm{d} x, $$ and the space $X$ of all measurable $u:\mathbb{R}^d \to \mathcal{H}$ with $||u||_X < \infty$ is a Banach space. If $\mathcal{H}$ is infinite-dimensional, then $X$ is strictly larger than $L^2(\mathbb{R}^d,\mathcal{H})$. Does anyone know whether the space $X$ has been analysed in the literature and can provide some references? I am particularly interested in the dual space of $X$.

How to define the average of a multivariable function in infinite space?

Posted: 15 Feb 2022 03:32 AM PST

We know that the average of a function $f(x)$ where $f: \mathbb{R} \rightarrow \mathbb{R}$ in the interval $[a,b]$ can be computed by $$ \langle f(x) \rangle =\frac{1}{b-a}\int_a^b f(x)dx. $$ How should we define the average of a function in an infinite domain, say $[a,+\infty)$ and $(-\infty, \infty)$, given that the limits $\lim_{x \rightarrow \pm \infty} f(x)$ exist? And how do we extend this to functions $f: \mathbb{R}^n \rightarrow \mathbb{R}$? Is it, for example, reasonable to say that the average of a function $f(\vec{x})$ is $$ \langle f(\vec{x}) \rangle = \lim_{L \rightarrow \infty} \frac{1}{L^n} \int_{-L}^L \int_{-L}^L \dots \int_{-L}^L f(\vec{x})d\vec{x} $$ ? What about choosing another coordinate system, say, a generalized spherical coordinate system instead, and integrate the function in a high-dimensional sphere with the radius approaching infinity? What properties does the function have to fulfill to ensure that all coordinate representations give equal results?

If any illustrative examples are used, I would prefer that a high-dimensional Gaussion function could be taken as an example, which is a perfect example of function that approaches to zero in all directions.

The problem that I am originally dealing with is to calculate the variance of the linear combination of Gaussian functions centered at different locations, which involves evaluating the average of something like this $$ \sum_{i,j} e^{-[(\vec{x}-\vec{x}_i)^2 + (\vec{x}-\vec{x}_j)^2]/2\sigma^2} $$ over the entire $\mathbb{R}^n$ space.

Is KL divergence between two discrete pdfs minimized when they are closest in Euclidean distance?

Posted: 15 Feb 2022 03:02 AM PST

I've been wondering about a simple question for few minutes, but I can't find any relevant questions or references by googling.

I want to minimize KL divergence from $p\in \Delta(A)$ where A is some discrete set to compact set $\Theta\subset \Delta(A)$ where $p \notin \Theta$. That is, I want to solve $$\min_{q\in\Theta}D(p||q)$$ Since $A$ is a discrete set, $\Theta, \Delta(A)$ is just a subset of $\mathbb{R}^{|A|}$.

Intuitively, since KL divergence measures "distance", I've conjectured that this would be minimized with $q$ closest to $p$ in Euclidean distance.

I'm not sure if I can prove it, or even if it is right at all. Does the following hold?

$$arg\min_{q\in\Theta}D(p||q)=arg\min_{q\in\Theta}d(p,q)$$ $d(p,q)$ is the standard Euclidean metric.

Any answers/counter examples/references would be helpful. Thank you!

The heat equation definition

Posted: 15 Feb 2022 03:10 AM PST

I was just running through some youtube videos on the Heat equation. In the heat equation shown in the image, why does the derivative of $T$ wrt to $t$ and second derivative wrt to $x$ have the same function of $x$?

In the formation of the heat equation second derivative term. The second term is derived from calculating the difference of gradients at sequential points on the function $T(x)$. So consider three points $x_1, x_2, x_3$. The change in temperature of $x_2$ wrt $t$ is proportional to $(x_1+x_3/2) - x_2$. Or $\partial_t T = a[(x_1+x_3/2) - x_2]$. $a[(x_1+x_3/2) - x_2]$ is equivalent to taking the second derivative of $T(x)$ and substituting the point $x_1$. Since $(x_1+x_3/2) - x_2$ is essentially equal to the difference between $(1/2)([x_3-x_2]-[x_2-x_1])$. But from the heat equation it shows $\partial_t T(x, t) = \partial_{xx}T(x,t)$, shouldn't it be $\partial_T T(x+1,t) = \partial_{xx} T(x, t)$ to show that the change in temperature of a given point $x_2$ wrt time is dependent on the second derivative value of $T$ wrt $x$ at the point $x_1$ instead of $x_2$?

Formation of heat equation

Heat equation

Project a parabolic curve to straight lines.

Posted: 15 Feb 2022 03:53 AM PST

I am working with Voronoi Diagrams and I'm given a parabolic curve that is described through four parameters:

  • A start point, say $P_0(x_0,y_0)$.
  • An end point, say $P_1(x_1,y_1)$.
  • A focus point, say $F(x_f,y_f)$.
  • And a directrix line $l$, say from point $L_0$ to $L_1$.

enter image description here

Instead of calculating the parabola points (as in the Wiki), I'd like to find a third point, let's say $C$, that creates two straight lines (e.g. horizontal and vertical) between $P_0$ and $P_1$, such that it "follows" the curve. For example, for the following image, if $P_0 = (3,3)$ and $P_1 = (5,1)$, then I want to find $C = (3,1)$.

I was looking into Bezier curves, but it does not seem to fulfill my needs. The control point seems to depend on the curvature of the parabola, thus it might not give the straight lines I seek.

Is there a proper way to calculate such a point?

enter image description here

EDIT: For the figure above there are two possible solutions to connect $P_0$ to $P_1$ with two straight lines (i.e. horizontal or vertical). Either we go east then south; or south then east. Following the curve means the later solution is the one I am seeking because it somewhat follows the parabola.

enter image description here

I could say point $C$ is farthest from the focal point; but rather than this simple algorithm I would like to know if there's a more robust and mathematical definition for such point, such as Bezier Curves. You could this simply does not exist and that would be a valid solution.

enter image description here

Brainstorming: I've thought about using the tangent crossing (as in the comments below) but they don't yield the exact point I need as they depend on the slope. I've also thought about interpreting this parabola as part of a circle and computing the tangent crossings, but, intuitively, that seems to run into the same problem.

Bound the norm of a matrix function related to discrete algebraic Riccati equation

Posted: 15 Feb 2022 03:35 AM PST

I was going through the following paper on perturbation analysis of the discrete Riccati equation. https://dml.cz/bitstream/handle/10338.dmlcz/124552/Kybernetika_29-1993-1_2.pdf.

The perturbation analysis involves the norm of the inverse of $\mathcal{T}:\mathbb{R}^{n\times n} \rightarrow \mathbb{R}^{n\times n}$. To be more specific, $\mathcal{T}(X)= X + \gamma A_c' XA_c$ with $0<\gamma\leq 1$, where the eigenvalues of $A_c\in\mathbb{R}^{n\times n}$ (i.e., $\lambda_i,i=1,2,\cdots,n$) lies inside the unique circle in the complex plane. That is $A_c$ is stable (discrete-time). Let $\mathcal{L}(\mathbb{R}^{n\times n},\mathbb{R}^{n\times n})$ be the space of linear operators $\mathbb{R}^{n\times n}\rightarrow \mathbb{R}^{n\times n}$ with the following induced norm $$ \Vert \mathcal{T}\Vert_\mathcal{L} = \max \{\Vert \mathcal{\mathcal{T}}(X)\Vert:\Vert X\Vert =1 \}, \ \ \ \textrm{for } \mathcal{T}\in \mathcal{L}(\mathbb{R}^{n\times n},\mathbb{R}^{n\times n}). $$

Note that the eigenvalues of $\mathcal{T}$ are $\mu_{ij}=1-\lambda_i\lambda_j$. Hence, $\mathcal{T}$ is invertible. We are interested in $\Vert \mathcal{T}^{-1}\Vert_{\mathcal{L}}$ or other well-defined norms of $\mathcal{T}^{-1}$.

Since we know $A_c$, I was wondering if we can write $\Vert \mathcal{T}^{-1} \Vert$ explicitly as a function of eigenvalues of $A_c$ or provide an upper bound on $\Vert \mathcal{T}^{-1} \Vert$ that depends on $A_c$.

My attempt:

Since $A_c$ is stable, there exists $L$ such that $L = X - \gamma A_c' X A_c$, which is a Lyapunov equation given $L$ and $A_c$. Then, $X = \sum_{m=0}^\infty \gamma^m (A_c')^m L (A_c)^m$. Hence, we have $\Vert \mathcal{T}^{-1}(L)\Vert = \Vert \sum_{m=0}^\infty \gamma^m (A_c')^m L (A_c)^m \Vert \leq \sum_{m=0}^\infty \gamma^m\Vert A_c^m\Vert^{2}\Vert L\Vert$. Then, $\Vert \mathcal{T}^{-1}\Vert\leq \sum_{m=0}^{\infty}\gamma^m \Vert A_c^m\Vert^{2}$.

The special radius of a matrix is bounded by a norm of the matrix. Can we apply this fact before the triangular inequality is applied and derive a tighter bound on $\Vert \mathcal{T}^{-1}\Vert$ using the eigenvalues of $A_c$?

Integration of $\int\sqrt{\frac{a+x}{a-x}} dx$, exercise $45$ in section $6.22$ of Tom Apostol's Calculus Vol $1$

Posted: 15 Feb 2022 03:47 AM PST

While looking at the solution here: https://www.stumblingrobot.com/2015/12/01/evaluate-the-integral-of-axa-x12/

Evaluate the following integral for $a>0$. $$\newcommand{\dd}{\; \mathrm{d}} \int\sqrt{\frac{a+x}{a-x}} \dd x.$$ First, we multiply the numerator and denominator in the fraction inside the square root by $a+x$ and do some simplification, \begin{align*} \int \sqrt{\frac{a+x}{a-x}} \, dx &= \int \sqrt{\frac{(a+x)^2}{(a-x)(a+x)}} \, dx \\[9pt] &= \int \frac{a+x}{\sqrt{a^2-x^2}} \, dx \\[9pt] &= \int \frac{a}{\sqrt{a^2-x^2}} \,dx + \int \frac{x}{\sqrt{a^2-x^2}} \, dx \\[9pt] &= \int \frac{1}{\sqrt{1 - \left( \frac{x}{a} \right)^2}} \, dx + \int \frac{x}{\sqrt{a^2-x^2}} \, dx. \end{align*} Now, we evaluate these two integrals separately. ...

I wondered why it is fine to multiply the numerator and denominator by $(a+x)$?

In the original integrand, the square root has to be non-negative and the denominator has to be non-0, so $x \in [-a, a)$. Therefore, the end point $-a$ is still in the domain of the integrand, but it is not in the domain of the new integrand when we extend it with $(a+x)/(a+x)$, so I'm not sure why exactly that is justified.

If that's justified, please explain why. If that's not justified, please provide a full correct solution.

Definition of rank of matrices and nullity of matrices

Posted: 15 Feb 2022 03:43 AM PST

I am getting confused with certain definitions regarding matrices. Suppose we have $(m,n)$ matrix, $A$ whose entries come from a PID $R$.

  1. Is the rank of the matrix $A$ defined to be the:

    i) maximal number of linearly independent columns or

    ii) is it the rank of the column space or

    iii) are they both the same?

  2. I know that the rank-nullity theorem applies to modules over an integral domain $R$.

i.e. if $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is exact then $$0\rightarrow A\otimes k \rightarrow B\otimes k\rightarrow C\otimes k\rightarrow 0$$ is exact where $k$ is the field of fractions of $R$. Hence,

$$ \mbox{rank} (B) = \mbox{rank} (B \otimes k) = \mbox{rank}(A) + \mbox{rank}(C).$$

But does this also apply to matrices in $R$? I suspect that the proof of rank-nullity theorem for matrices in wikipedia can only hold for fields.

What is the probability that if a randomly chosen product is defective, then it came from machine X?

Posted: 15 Feb 2022 03:34 AM PST

Two machines, A and B, each independently produces a product. Machine X makes 70 percent of the product while Y makes 30 percent. From past statistics, 5 percent made by X and 6 percent made by Y are defective. Given that a product randomly chosen was defective, what is the probability it came from X?

How could I start with this?

Thanks.

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